Question
Periodic Function Flexibility Practice
Complete this assessment to review what you’ve learned. It will not count toward your grade.
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Question
Use the image to answer the question.
A waveform is drawn on a coordinate plane. The x-axis ranges from start fraction negative pi over 2 end fraction to 2 pi in increments of start fraction pi over 4 end fraction. The y-axis ranges from negative 3 to 3 in unit increments.
Using the given graph, write an equation for a sine function.
(1 point)
y=
Complete this assessment to review what you’ve learned. It will not count toward your grade.
4 of 54 of 5 Items
Question
Use the image to answer the question.
A waveform is drawn on a coordinate plane. The x-axis ranges from start fraction negative pi over 2 end fraction to 2 pi in increments of start fraction pi over 4 end fraction. The y-axis ranges from negative 3 to 3 in unit increments.
Using the given graph, write an equation for a sine function.
(1 point)
y=
Answers
GPT-4o mini
To write the equation for a sine function based on the waveform you've described, we can use the general form of a sine function, which is:
\[ y = A \sin(B(x - C)) + D \]
Where:
- \( A \) is the amplitude (the height from the middle of the wave to its peak),
- \( B \) affects the period of the function,
- \( C \) is the phase shift (horizontal shift),
- \( D \) is the vertical shift.
From your description, let's analyze each parameter:
1. **Amplitude (A)**: If the wave oscillates between -3 and 3, the amplitude is:
\[
A = \frac{\text{Maximum} - \text{Minimum}}{2} = \frac{3 - (-3)}{2} = 3
\]
2. **Vertical Shift (D)**: The midline of the wave is at 0 since it oscillates around the horizontal line \( y = 0 \). Thus:
\[
D = 0
\]
3. **Period (B)**: To find the period, observe how long it takes for the wave to repeat itself. If the wave completes one full cycle from \( -\frac{\pi}{2} \) to \( \frac{3\pi}{2} \) (which is \( 2\pi \)), then:
\[
\text{Period} = \left(\text{Final x} - \text{Initial x}\right) = \frac{3\pi}{2} - (-\frac{\pi}{2}) = 2\pi
\]
Since the standard period of \( \sin(x) \) is \( 2\pi \), \( B = 1 \):
\[
B = \frac{2\pi}{\text{Period}} = \frac{2\pi}{2\pi} = 1
\]
4. **Phase Shift (C)**: If the sine wave starts at \( -\frac{\pi}{2} \) (which can be interpreted as a cosine function starting point), we note there is a shift. A sine wave that starts from its midline going upwards at \( 0 \) would have its shift:
\[
x \rightarrow \left(x + \frac{\pi}{2}\right)
\]
So in this case, \( C = -\frac{\pi}{2} \).
Putting it all together, we arrive at the function:
\[
y = 3 \sin\left(1\left(x + \frac{\pi}{2}\right)\right) + 0
\]
Thus, simplifying, the final function can be written as:
\[
y = 3 \sin\left(x + \frac{\pi}{2}\right)
\]
If needed in another form, remember \( \sin(x + \frac{\pi}{2}) = \cos(x) \), so:
\[
y = 3 \cos(x)
\]
However, since we are asked for a sine function, stick with:
\[
y = 3 \sin\left(x + \frac{\pi}{2}\right)
\]
So the answer is:
\( y = 3 \sin\left(x + \frac{\pi}{2}\right) \)
\[ y = A \sin(B(x - C)) + D \]
Where:
- \( A \) is the amplitude (the height from the middle of the wave to its peak),
- \( B \) affects the period of the function,
- \( C \) is the phase shift (horizontal shift),
- \( D \) is the vertical shift.
From your description, let's analyze each parameter:
1. **Amplitude (A)**: If the wave oscillates between -3 and 3, the amplitude is:
\[
A = \frac{\text{Maximum} - \text{Minimum}}{2} = \frac{3 - (-3)}{2} = 3
\]
2. **Vertical Shift (D)**: The midline of the wave is at 0 since it oscillates around the horizontal line \( y = 0 \). Thus:
\[
D = 0
\]
3. **Period (B)**: To find the period, observe how long it takes for the wave to repeat itself. If the wave completes one full cycle from \( -\frac{\pi}{2} \) to \( \frac{3\pi}{2} \) (which is \( 2\pi \)), then:
\[
\text{Period} = \left(\text{Final x} - \text{Initial x}\right) = \frac{3\pi}{2} - (-\frac{\pi}{2}) = 2\pi
\]
Since the standard period of \( \sin(x) \) is \( 2\pi \), \( B = 1 \):
\[
B = \frac{2\pi}{\text{Period}} = \frac{2\pi}{2\pi} = 1
\]
4. **Phase Shift (C)**: If the sine wave starts at \( -\frac{\pi}{2} \) (which can be interpreted as a cosine function starting point), we note there is a shift. A sine wave that starts from its midline going upwards at \( 0 \) would have its shift:
\[
x \rightarrow \left(x + \frac{\pi}{2}\right)
\]
So in this case, \( C = -\frac{\pi}{2} \).
Putting it all together, we arrive at the function:
\[
y = 3 \sin\left(1\left(x + \frac{\pi}{2}\right)\right) + 0
\]
Thus, simplifying, the final function can be written as:
\[
y = 3 \sin\left(x + \frac{\pi}{2}\right)
\]
If needed in another form, remember \( \sin(x + \frac{\pi}{2}) = \cos(x) \), so:
\[
y = 3 \cos(x)
\]
However, since we are asked for a sine function, stick with:
\[
y = 3 \sin\left(x + \frac{\pi}{2}\right)
\]
So the answer is:
\( y = 3 \sin\left(x + \frac{\pi}{2}\right) \)